What is the integral of 1-(x^2)/4 ?

The integral of 1-(x^2)/4 is x-(x^3)/(12)+C

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integral of 1-(x^2)/4

\int 1 - \frac{x ^{2}}{4} d x

x - \frac{x ^{3}}{12} + C

Solution steps

\int 1 - \frac{x ^{2}}{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int \frac{x ^{2}}{4} d x

\int 1 d x = x

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

= x - \frac{x ^{3}}{12}

Add a constant to the solution

= x - \frac{x ^{3}}{12} + C

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